6
$\begingroup$

Boiling point of ammonia is $-33.35\ \mathrm{^{\circ}C}$ and its critical temperature $132.4\ \mathrm{^{\circ}C}$ is well above the room temperature. So we can liquify the ammonia gas by compressing it into a bottle, and store it at room temperature for ever.
(Even though the bottle is kept at room temperature, the ammonia will stay in liquid state due to high pressure. )

Hope I'm correct with the above thinking. Next, I open the bottle and pour some liquid ammonia into a bowl. Since this bowl is at 1 atm, I expect something to happen. Does the temperature suddenly drop to $-33.35\ \mathrm{^{\circ}C}$? If so, why?

I'm trying to relate this to adiabatic expansion of gas – the gas does work on the surrounding as it expands and loses its kinetic energy. But here ammonia is in liquid state as its temperature drops to $-33.35\ \mathrm{^{\circ}C}$. I don't see any work getting done by the ammonia, so I don't see how the temperature drops. I feel I'm missing some important concept.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ When the pressure drops the ammonia wants to evaporate. but evaporation costs energy which can only come from the environment (which consists of the bowl and the ammonia). Now do you see what happens? $\endgroup$
    – matt_black
    Mar 18, 2017 at 16:50
  • $\begingroup$ At 1 atm and room temperature, the ammonia in the bowl is well above its boiling point. So yeah it does want to evaporate. Does the energy required for the evaporation come from the liquid ammonia itself ? Thus, cooling the liquid ! Wow! Thank you (: $\endgroup$
    – Hiiii
    Mar 18, 2017 at 17:00
  • $\begingroup$ Hey just a small question. Can the liquid ammonia cool below its boiling point due to evaporation ? I know that the temperature stays constant during state change. But it seems here the situation is different, I don't see why it cannot go below the boiling point.. $\endgroup$
    – Hiiii
    Mar 18, 2017 at 17:07
  • 2
    $\begingroup$ The temperature can go pretty cold. The situation is not in equilibrium so the simple idea that the temperature stays the same in a state change doesn't apply. $\endgroup$
    – matt_black
    Mar 18, 2017 at 17:13
  • 2
    $\begingroup$ @Hiiii liquids want to evaporate because the vapor pressure in the surrounding air is less than the saturation vapor pressure, not because of some relation to boiling temperature. The boiling temperature is just the case where saturation vapor pressure equals the ambient atmospheric pressure. $\endgroup$
    – casey
    Mar 18, 2017 at 20:44

1 Answer 1

Reset to default
11
$\begingroup$

The key to understanding what happens is that evaporation costs energy.

Changing the state of a liquid to a gas requires the input of energy: this is basic thermodynamics. This is in addition to any effects related to a change in temperature which also costs energy or releases energy.

When liquid ammonia evaporates to gaseous ammonia, energy is required from somewhere. If liquid ammonia at room temperature is poured into a bowl at room temperature it will want to evaporate as it is above its boiling point. But that evaporation requires an input of energy. That energy will come from the environment which consists of the bowl and the rest of the liquid. The only way to provide that energy is to cool the rest of the liquid and the bowl. This happens quickly and the temperature will often fall locally to below the boiling point of ammonia as there isn't enough time to equilibrate the temperature of the liquid with the temperature of the environment surrounding the bowl.

The net effect is that you will end up with some liquid ammonia at some temperature below its boiling point and a lot of gaseous ammonia (don't do this outside a good fume hood).

Forced evaporation of volatile liquids used to be used to freeze water via a similar process. For example, place a small bowl of water in a larger bowl of ether. Blow air onto the ether, which will evaporate, drawing energy from its environment (including the remaining ether and the bowl of water). The water will freeze if you used enough ether. I don't recommend doing this as ether is very flammable, but it was once a common demonstration of this process in chemistry classes.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Correction: the liquid ammonia will boil until it reaches its boiling point and no further. It will not drop below its boiling point. Also, the process will happen essentially instantaneously, i.e. in the valve where the pressure is released. $\endgroup$
    – Level River St
    Mar 18, 2017 at 18:29
  • $\begingroup$ @LevelRiverSt You are only right if the mixture is at equilibrium. Otherwise it is easy to get colder than the boiling point. $\endgroup$
    – matt_black
    Mar 18, 2017 at 18:33
  • 1
    $\begingroup$ Why do I miss here any consideration of the Joule-Thomson effect here? With the hint in the original question of a critical temperature of Ammonia of 132 degC, and the rule that the Joule-Thomson inversion temperature of a van der Waals-like Gas of 6.75 times the critical temperature (in K), any sample of ammonia gas initially stored at room temperature and under pressure will cool already by the mere expansion. If the sample is just large enough, and the system is kept adiabatic... $\endgroup$
    – Buttonwood
    Mar 19, 2017 at 0:16
  • $\begingroup$ @Buttonwood It would be interesting to see if that is a significant contributor to cooling in this circumstance. It might not be as the expansion is not happening in an insulated system with no heat transfer. $\endgroup$
    – matt_black
    Mar 19, 2017 at 0:24
  • 1
    $\begingroup$ @matt_black I agree it were interesting because the reasoning assumes the sample were already a gas, and not a liquid. To point to Joule-Thomson effect simply came to my mind as recently I read about ammonia used as a coolant for a bobsleigh run, that would allow to use the track even if the ambient temperature would settle around 20 degC (45 t of ammonia in the circuit and four 250 kW aggregates...) $\endgroup$
    – Buttonwood
    Mar 19, 2017 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.